Question

The tangent plane to the surface \( z = \sqrt{x^2 + 3y^2} \) at (1, 1, 2) is given by

• A. \( x - 3y + z = 0 \)
• B. \( x + 3y - 2z = 0 \)
• C. \( 2x + 4y - 3z = 0 \)
• D. \( 3x - 7y + 2z = 0 \)

Hint:

To find the equation of the tangent plane, use the partial derivatives of the surface function with respect to \( x \) and \( y \) and plug them into the tangent plane formula.

Answer 1

The Correct Option is B

Step 1: Tangent Plane Formula. 
The equation for the tangent plane to a surface at a point \( (x_0, y_0, z_0) \) is given by: \[ z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0)(y - y_0) \] We are given the surface \( z = \sqrt{x^2 + 3y^2} \). At the point (1, 1, 2), we need to compute the partial derivatives of \( z \) with respect to \( x \) and \( y \). 
 

Step 2: Computing Partial Derivatives. 
We compute: \[ \frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 + 3y^2}}, \frac{\partial z}{\partial y} = \frac{3y}{\sqrt{x^2 + 3y^2}} \] At the point (1, 1, 2), these partial derivatives become: \[ \frac{\partial z}{\partial x}(1, 1) = \frac{1}{\sqrt{1^2 + 3(1)^2}} = \frac{1}{2}, \frac{\partial z}{\partial y}(1, 1) = \frac{3}{\sqrt{1^2 + 3(1)^2}} = \frac{3}{2} \] Thus, the equation for the tangent plane is: \[ z - 2 = \frac{1}{2}(x - 1) + \frac{3}{2}(y - 1) \] Simplifying: \[ z = 2 + \frac{1}{2}(x - 1) + \frac{3}{2}(y - 1) \] \[ x + 3y - 2z = 0 \]

Step 3: Conclusion. 
The correct equation for the tangent plane is (B) \( x + 3y - 2z = 0 \).